376 Wiggle Subsequence 摆动序列
Description:
A sequence of numbers is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.
For example, [1,7,4,9,2,5] is a wiggle sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, [1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.
Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
Example:
Example 1:
Input: [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.
Example 2:
Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
Example 3:
Input: [1,2,3,4,5,6,7,8,9]
Output: 2
Follow up:
Can you do it in O(n) time?
题目描述:
如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差（如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。
例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值都是正数，第二个序列是因为它的最后一个差值为零。
给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些（也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。
示例 :
示例 1:
输入: [1,7,4,9,2,5]
输出: 6
解释: 整个序列均为摆动序列。
示例 2:
输入: [1,17,5,10,13,15,10,5,16,8]
输出: 7
解释: 这个序列包含几个长度为 7 摆动序列，其中一个可为[1,17,10,13,10,16,8]。
示例 3:
输入: [1,2,3,4,5,6,7,8,9]
输出: 2
进阶:
你能否用 O(n) 时间复杂度完成此题?
思路:
动态规划
dp[i]表示 nums前 i个数中的摆动序列
如果 nums[i]和前后两个数一正一负, 则 dp[i] = dp[i - 1] + 1
可以优化为两个常数空间
dp[i][0] 表示前 i个数中的摆动序列, 且 nums[i] > nums[i - 1]
dp[i][1] 表示前 i个数中的摆动序列, 且 nums[i] < nums[i - 1]
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int wiggleMaxLength(vector<int>& nums)
{
int up = nums.empty() ? 0 : 1, down = nums.empty() ? 0 : 1;
for (int i = 1; i < nums.size(); i++)
{
if (nums[i] > nums[i - 1]) up = down + 1;
if (nums[i] < nums[i - 1]) down = up + 1;
}
return max(up, down);
}
};
Java:
class Solution {
public int wiggleMaxLength(int[] nums) {
int up = nums.length == 0 ? 0 : 1, down = nums.length == 0 ? 0 : 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) up = down + 1;
if (nums[i] < nums[i - 1]) down = up + 1;
}
return Math.max(up, down);
}
}
Python:
class Solution:
def wiggleMaxLength(self, nums: List[int]) -> int:
up, down = 1 if nums else 0, 1 if nums else 0
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
up = down + 1
if nums[i] < nums[i - 1]:
down = up + 1
return max(up, down)
